Ed Slides Actual

7/29/2019 Ed Slides Actual

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Contents

1. Scales

2. Orthographic Projections - Basics

3. Conversion of Pictorial View into Orthographic Views

4. Projections of Points and Lines

5. Projection of Planes

6. Projection of Solids

EXIT

7. Sections & Development8. Intersection of Surfaces

9. Isometric Projections

Exercises

Solutions Applications of Lines


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Scales

1. Basic Information

2. Types and important units

3. Plain Scales (3 Problems)

4. Diagonal Scales - information

5. Diagonal Scales (3 Problems)


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Orthographic Projections - Basics

1. Drawing The fact about

2. Drawings - Types

3. Orthographic (Definitions and Important terms)

4. Planes - Classifications

5. Pattern of planes & views

6. Methods of orthographic projections

7. 1st

angle and 3rd

angle method two illustrations


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Conversion of pictorial views in to orthographic views.

1. Explanation of various terms

2. 1st angle method - illustration

3. 3rd angle method illustration

4. To recognize colored surfaces and to draw three Views

5. Seven illustrations (no.1 to 7) draw different orthographic views

6. Total nineteen illustrations ( no.8 to 26)


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Projection of Points and Lines

1. Projections Information

2. Notations

3. Quadrant Structure.

5. Projections of a Point in 1st quadrant.

6. Lines Objective & Types.

7. Simple Cases of Lines.

8. Lines inclined to one plane.

9. Lines inclined to both planes.

10. Imp. Observations for solution11. Important Diagram & Tips.

12. Group A problems 1 to 5

13. Traces of Line ( HT & VT )

14. To locate Traces.

15. Group B problems: No. 6 to 8

16. HT-VT additional information.

17. Group B1 problems: No. 9 to 11

18. Group B1 problems: No. 9 to 1

4. Object in different Quadrants Effect on position of views.

19. Lines in profile plane

20. Group C problems: No.12 & 13

21. Applications of Lines:: Information

22. Group D: Application Problems: 14 to 23


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Projections of Planes:

1. About the topic:

2. Illustration of surface & side inclination.

3. Procedure to solve problem & tips:

4. Problems:1 to 5: Direct inclinations:

5. Problems:6 to 11: Indirect inclinations:

6. Freely suspended cases: Info:

7. Problems: 12 & 13

8. Determination of True Shape: Info:

9. Problems: 14 to 17


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Projections of Solids:1. Classification of Solids:

2. Important parameters:

3. Positions with Hp & Vp: Info:

4. Pattern of Standard Solution.

5. Problem no 1,2,3,4: General cases:

6. Problem no 5 & 6 (cube & tetrahedron)

7. Problem no 7 : Freely suspended:

8. Problem no 8 : Side view case:

9. Problem no 9 : True length case:

10. Problem no 10 & 11 Composite solids:

11. Problem no 12 : Frustum & auxiliary plane:


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Section & Development1. Applications of solids:

2. Sectioning a solid: Information:

3. Sectioning a solid: Illustration Terms:

4. Typical shapes of sections & planes:

5. Development: Information:

6. Development of diff. solids:

7. Development of Frustums:

8. Problems: Standing Prism & Cone: no. 1 & 2

9. Problems: Lying Prism & Cone: no.3 & 4

10. Problem: Composite Solid no. 5

11. Problem: Typical cases no.6 to 9


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Intersection of Surfaces:1. Essential Information:

2. Display of Engineering Applications:

3. Solution Steps to solve Problem:

4. Case 1: Cylinder to Cylinder:

5. Case 2: Prism to Cylinder:

6. Case 3: Cone to Cylinder

7. Case 4: Prism to Prism: Axis Intersecting.

8. Case 5: Triangular Prism to Cylinder

9. Case 6: Prism to Prism: Axis Skew

10. Case 7 Prism to Cone: from top:

11. Case 8: Cylinder to Cone:


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Isometric Projections

1. Definitions and explanation

2. Important Terms

3. Types.

4. Isometric of plain shapes-1.

5. Isometric of circle

6. Isometric of a part of circle

7. Isometric of plain shapes-2

8. Isometric of solids & frustums (no.5 to 16)

9. Isometric of sphere & hemi-sphere (no.17 & 18)

10. Isometric of Section of solid.(no.19)

11. Illustrated nineteen Problem (no.20 to 38)


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FOR FULL SIZE SCALER.F.=1 OR ( 1:1 )

MEANS DRAWING& OBJECT ARE OFSAME SIZE.

Other RFs are describedas

1:10, 1:100,1:1000, 1:1,00,000

SCALES

DIMENSIONS OF LARGE OBJECTS MUST BE REDUCED TO ACCOMMODATEON STANDARD SIZE DRAWING SHEET.THIS REDUCTION CREATES A SCALE

OF THAT REDUCTION RATIO, WHICH IS GENERALLY A FRACTION..

SUCH A SCALE IS CALLED REDUCING SCALEAND

THAT RATIO IS CALLED REPRESENTATIVE FACTOR.

SIMILARLY IN CASE OF TINY OBJECTS DIMENSIONS MUST BE INCREASEDFOR ABOVE PURPOSE. HENCE THIS SCALE IS CALLED ENLARGING SCALE.HERE THE RATIO CALLED REPRESENTATIVE FACTOR IS MORE THAN UNITY.

REPRESENTATIVE FACTOR (R.F.)=

=

=

=

A

USE FOLLOWING FORMULAS FOR THE CALCULATIONS IN THIS TOPIC.

B LENGTH OF SCALE =R.F. MAX. LENGTH TO BE MEASURED.X

DIMENSION OF DRAWING

DIMENSION OF OBJECT

LENGTH OF DRAWING

ACTUAL LENGTH

AREA OF DRAWING

ACTUAL AREA

VOLUME AS PER DRWG.

ACTUAL VOLUME

V

V3


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1. PLAIN SCALES ( FOR DIMENSIONS UP TO SINGLE DECIMAL)2. DIAGONAL SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS)

3. VERNIER SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS)4. COMPARATIVE SCALES ( FOR COMPARING TWO DIFFERENT UNITS)5. SCALE OF CORDS ( FOR MEASURING/CONSTRUCTING ANGLES)

TYPES OF SCALES:

= 10 HECTOMETRES= 10 DECAMETRES= 10 METRES= 10 DECIMETRES= 10 CENTIMETRES= 10 MILIMETRES

1 KILOMETRE1 HECTOMETRE1 DECAMETRE1 METRE1 DECIMETRE1 CENTIMETRE

BE FRIENDLY WITH THESE UNITS.


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0 1 2 3 4 510

PLAIN SCALE:-This type of scale represents two units or a unit and its sub-division.

METERS

DECIMETERSR.F. = 1/100

4 M 6 DM

PLANE SCALE SHOWING METERS AND DECIMETERS.

PLAIN SCALE

PROBLEM NO.1:- Draw a scale 1 cm = 1m to read decimeters, to measure maximum distance of 6 m.

Show on it a distance of 4 m and 6 dm.

CONSTRUCTION:-

a) Calculate R.F.=

R.F.= 1cm/ 1m = 1/100

Length of scale = R.F. X max. distance

= 1/100 X 600 cm

= 6 cms

b) Draw a line 6 cm long and divide it in 6 equal parts. Each part will represent larger division unit.

c) Sub divide the first part which will represent second unit or fraction of first unit.

d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisionson left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale.

e) After construction of scale mention its RF and name of scale as shown.

f) Show the distance 4 m 6 dm on it as shown.

DIMENSION OF DRAWING

DIMENSION OF OBJECT


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PROBLEM NO.2:- In a map a 36 km distance is shown by a line 45 cms long. Calculate the R.F. and constructa plain scale to read kilometers and hectometers, for max. 12 km. Show a distance of 8.3 km on it.

CONSTRUCTION:-a) Calculate R.F.

R.F.= 45 cm/ 36 km = 45/ 36 . 1000 . 100 = 1/ 80,000Length of scale = R.F. max. distance

= 1/ 80000 12 km

= 15 cm

b) Draw a line 15 cm long and divide it in 12 equal parts. Each part will represent larger division unit.

c) Sub divide the first part which will represent second unit or fraction of first unit.

d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions

on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale.

e) After construction of scale mention its RF and name of scale as shown.

f) Show the distance 8.3 km on it as shown.

KILOMETERSHECTOMETERS

8KM 3HM

R.F. = 1/80,000PLANE SCALE SHOWING KILOMETERS AND HECTOMETERS

0 1 2 3 4 5 6 7 8 9 10 1110 5

PLAIN SCALE


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PROBLEM NO.3:- The distance between two stations is 210 km. A passenger train covers this distancein 7 hours. Construct a plain scale to measure time up to a single minute. RF is 1/200,000 Indicate the distance

traveled by train in 29 minutes.

CONSTRUCTION:-a) 210 km in 7 hours. Means speed of the train is 30 km per hour ( 60 minutes)

Length of scale = R.F. max. distance per hour

= 1/ 2,00,000 30km

= 15 cm

b) 15 cm length will represent 30 km and 1 hour i.e. 60 minutes.

Draw a line 15 cm long and divide it in 6 equal parts. Each part will represent 5 km and 10 minutes.

c) Sub divide the first part in 10 equal parts,which will represent second unit or fraction of first unit.

Each smaller part will represent distance traveled in one minute.

d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisionson left-hand side of Zero. Take height of scale 5 to 10 mm for getting a proper look of scale.

e) Show km on upper side and time in minutes on lower side of the scale as shown.

After construction of scale mention its RF and name of scale as shown.

f) Show the distance traveled in 29 minutes, which is 14.5 km, on it as shown.

PLAIN SCALE

0 10 20 30 40 5010 MINUTESMIN

R.F. = 1/100PLANE SCALE SHOWING METERS AND DECIMETERS.

KMKM 0 5 10 15 20 255 2.5

DISTANCE TRAVELED IN 29 MINUTES.

14.5 KM


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We have seen that the plain scales give only two dimensions,such as a unit and its subunit or its fraction.

1

2

3

4

56

7

8

9

10X

Y

Z

The principle of construction of a diagonal scale is as follows.Let the XY in figure be a subunit.

From Y draw a perpendicular YZ to a suitable height.

Join XZ. Divide YZ in to 10 equal parts.

Draw parallel lines to XY from all these divisions

and number them as shown.

From geometry we know that similar triangles have

their like sides proportional.

Consider two similar triangles XYZ and 7 7Z,

we have 7Z / YZ = 77 / XY (each part being one unit)

Means 7 7 = 7 / 10. x X Y = 0.7 XY

:.

Similarly

1 1 = 0.1 XY

2 2 = 0.2 XYThus, it is very clear that, the sides of small triangles,

which are parallel to divided lines, become progressively

shorter in length by 0.1 XY.

The solved examples ON NEXT PAGES will

make the principles of diagonal scales clear.

The diagonal scales give us three successive dimensionsthat is a unit, a subunit and a subdivision of a subunit.

DIAGONAL

SCALE


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R.F. = 1 / 40,00,000

DIAGONAL SCALE SHOWING KILOMETERS.

0 100 200 300 400 500100 50

109876543210

KMKM

KM

569 km

459 km

336 km

222 km

PROBLEM NO. 4 : The distance between Delhi and Agra is 200 km.

In a railway map it is represented by a line 5 cm long. Find its R.F.

Draw a diagonal scale to show single km. And maximum 600 km.

Indicate on it following distances. 1) 222 km 2) 336 km 3) 459 km 4) 569 km

SOLUTION STEPS: RF = 5 cm / 200 km = 1 / 40, 00, 000

Length of scale = 1 / 40, 00, 000 X 600 X 105

= 15 cm

Draw a line 15 cm long. It will represent 600 km.Divide it in six equal parts.( each will represent 100 km.)

Divide first division in ten equal parts.Each will represent 10 km.Draw a line upward from left end and

mark 10 parts on it of any distance. Name those parts 0 to 10 as shown.Join 9th sub-division of horizontal scale

with 10th division of the vertical divisions. Then draw parallel lines to this line from remaining sub divisions and

complete diagonal scale.

DIAGONAL

SCALE

PROBLEM NO 5 A t l l t f l d i 1 28 h t i t d b i il t l


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PROBLEM NO.5: A rectangular plot of land measuring 1.28 hectors is represented on a map by a similar rectangle

of 8 sq. cm. Calculate RF of the scale. Draw a diagonal scale to read single meter. Show a distance of 438 m on it.

Draw a line 15 cm long.

It will represent 600 m.Divide it in six equal parts.( each will represent 100 m.)

Divide first division in ten equal parts.Each will

represent 10 m.

Draw a line upward from left end and

mark 10 parts on it of any distance.

Name those parts 0 to 10 as shown.Join 9th sub-division

of horizontal scale with 10th

division of the vertical divisions.Then draw parallel lines to this line from remaining sub divisions

and complete diagonal scale.

DIAGONAL

SCALESOLUTION :

1 hector = 10, 000 sq. meters

1.28 hectors = 1.28 X 10, 000 sq. meters

= 1.28 X 104

X 104

sq. cm8 sq. cm area on map represents

= 1.28 X 104 X 104 sq. cm on land

1 cm sq. on map represents

= 1.28 X 10 4 X 104/ 8 sq cm on land

1 cm on map represent

= 1.28 X 10 4 X 104/ 8 cm

= 4, 000 cm1 cm on drawing represent 4, 000 cm, Means RF = 1 / 4000

Assuming length of scale 15 cm, it will represent 600 m.

0 100 200 300 400 500100 50

109876543210

M

M

M

438 meters

R.F. = 1 / 4000

DIAGONAL SCALE SHOWING METERS.


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109876

543210

CENTIMETRES

M

M

CM

R.F. = 1 / 2.5

DIAGONAL SCALE SHOWING CENTIMETERS.

0 5 10 155 4 3 2 1

PROBLEM NO.6:. Draw a diagonal scale of R.F. 1: 2.5, showing centimeters

and millimeters and long enough to measure up to 20 centimeters.

SOLUTION STEPS:R.F. = 1 / 2.5

Length of scale = 1 / 2.5 X 20 cm.

= 8 cm.1.Draw a line 8 cm long and divide it in to 4 equal parts.

(Each part will represent a length of 5 cm.)

2.Divide the first part into 5 equal divisions.

(Each will show 1 cm.)

3.At the left hand end of the line, draw a vertical line and

on it step-off 10 equal divisions of any length.

4.Complete the scale as explained in previous problems.

Show the distance 13.4 cm on it.

13 .4 CM

DIAGONAL

SCALE


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( A Graphical Representation)

The Fact about:If compared with Verbal or Written Description,

Drawings offer far better idea about the Shape, Size & Appearance ofany object or situation or location, that too in quite a less time.

Hence it has become the Best Media of Communication

not only in Engineering but in almost all Fields.


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Drawings(Some Types)

Nature Drawings

( landscape,

scenery etc.)Geographical

Drawings

( maps etc.)

Botanical Drawings

( plants, flowers etc.)

Zoological Drawings

(creatures, animals etc.)

Portraits

( human faces,expressions etc.)

Engineering Drawings,

(projections.)

Machine component DrawingsBuilding Related Drawings.

Orthographic Projections(Fv,Tv & Sv.-Mech.Engg terms)(Plan, Elevation- Civil Engg.terms)

(Working Drawings 2-D type)

Isometric ( Mech.Engg.Term.)or Perspective(Civil Engg.Term)(Actual Object Drawing 3-D)


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ORTHOGRAPHIC PROJECTIONS:

Horizontal Plane (HP),

Vertical Frontal Plane ( VP )

Side Or Profile Plane ( PP)

Planes.Pattern of planes & Pattern of viewsMethods of drawing Orthographic Projections

Different Reference planes are

FV is a view projected on VP.

TV is a view projected on HP.

SV is a view projected on PP.

AndDifferent Views are Front View (FV), Top View (TV) and Side View (SV)

IMPORTANT TERMS OF ORTHOGRAPHIC PROJECTIONS:

IT IS A TECHNICAL DRAWING IN WHICH DIFFERENT VIEWS OF AN OBJECT

ARE PROJECTED ON DIFFERENT REFERENCE PLANES

OBSERVING PERPENDICULAR TO RESPECTIVE REFERENCE PLANE

123


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A.V.P.

to Hp & to Vp

PLANES

PRINCIPAL PLANES

HP AND VP

AUXILIARY PLANES

Auxiliary Vertical Plane

(A.V.P.)Profile Plane

( P.P.)

Auxiliary Inclined Plane

(A.I.P.)

1

PATTERN OF PLANES & VIEWS (First Angle Method)


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THIS IS A PICTORIAL SET-UP OF ALL THREE PLANES.

ARROW DIRECTION IS A NORMAL WAY OF OBSERVING THE OBJECT.

BUT IN THIS DIRECTION ONLY VP AND A VIEW ON IT (FV) CAN BE SEEN.

THE OTHER PLANES AND VIEWS ON THOSE CAN NOT BE SEEN.

HP IS ROTATED DOWNWARD 900

AND

BROUGHT IN THE PLANE OF VP.

PP IS ROTATED IN RIGHT SIDE 90

0

AND

BROUGHT IN THE PLANE OF VP.

X

Y

X Y

VP

HP

PP

FV

ACTUAL PATTERN OF PLANES & VIEWS

OF ORTHOGRAPHIC PROJECTIONS

DRAWN IN

FIRST ANGLE METHOD OF PROJECTIONS

LSV

TV

PROCEDURE TO SOLVE ABOVE PROBLEM:-

TO MAKE THOSE PLANES ALSO VISIBLE FROM THE ARROW DIRECTION,A) HP IS ROTATED 900 DOUNWARD

B) PP, 900 IN RIGHT SIDE DIRECTION.

THIS WAY BOTH PLANES ARE BROUGHT IN THE SAME PLANE CONTAINING VP.

PATTERN OF PLANES & VIEWS (First Angle Method)

2

Click to view Animation On clicking the button if a warning comes please click YES to continue, this program is

safe for your pc.
http://localhost/var/www/apps/conversion/tmp/scratch_2/PicView.exehttp://localhost/var/www/apps/conversion/tmp/scratch_2/PicView.exe


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Methods of Drawing Orthographic Projections

First Angle Projections MethodHere views are drawn

by placing objectin 1st Quadrant

( Fv above X-y, Tv below X-y )

Third Angle Projections MethodHere views are drawn

by placing objectin 3rd Quadrant.

( Tv above X-y, Fv below X-y)

FV

TV

X Y X Y

G L

TV

FV

SYMBOLIC

PRESENTATION

OF BOTH METHODS

WITH AN OBJECT

STANDING ON HP ( GROUND)ON ITS BASE.

3

NOTE:-HP term is used in 1st Angle method

&For the same

Ground term is usedin 3rd Angle method of projections

FIRST ANGLE


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FOR T.V.FIRST ANGLEPROJECTION

IN THIS METHOD,THE OBJECT IS ASSUMED TO BE

SITUATED IN FIRST QUADRANTMEANSABOVE HP & INFRONT OF VP.

OBJECT IS INBETWEENOBSERVER & PLANE.

ACTUAL PATTERN OFPLANES & VIEWS

IN

FIRST ANGLE METHODOF PROJECTIONS

X Y

VP

HP

PP

FV LSV

TV

G


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FOR T.V.


SITUATED IN THIRD QUADRANT( BELOW HP & BEHIND OF VP. )

PLANES BEING TRANSPERENTAND INBETWEEN

OBSERVER & OBJECT.


OFTHIRD ANGLE PROJECTIONS

X Y

TV

THIRD ANGLEPROJECTION

LSV FV


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ORTHOGRAPHIC PROJECTIONS{ MACHINE ELEMENTS }

OBJECT IS OBSERVED IN THREE DIRECTIONS.

THE DIRECTIONS SHOULD BE NORMAL

TO THE RESPECTIVE PLANES.

AND NOW PROJECT THREE DIFFERENT VIEWS ON THOSE PLANES.

THESE VEWS ARE FRONT VIEW , TOP VIEW AND SIDE VIEW.

FRONT VIEW IS A VIEW PROJECTED ON VERTICAL PLANE ( VP )

TOP VIEW IS A VIEW PROJECTED ON HORIZONTAL PLANE ( HP )

SIDE VIEW IS A VIEW PROJECTED ON PROFILE PLANE ( PP )

FIRST ANGLE


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FOR T.V.FIRST ANGLEPROJECTION


SITUATED IN FIRST QUADRANTMEANSABOVE HP & INFRONT OF VP.

OBJECT IS INBETWEENOBSERVER & PLANE.


IN

FIRST ANGLE METHODOF PROJECTIONS

X Y

VP

HP

PP

FV LSV

TV

FOR T.V.THIRD ANGLE


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OFTHIRD ANGLE PROJECTIONS

X

TV

LSV FV


SITUATED IN THIRD QUADRANT( BELOW HP & BEHIND OF VP. )

PLANES BEING TRANSPERENTAND INBETWEEN

OBSERVER & OBJECT.

FOR T.V.

Y

THIRD ANGLEPROJECTION


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x y

FRONT VIEW

TOP VIEW

L.H.SIDE VIEW

FOR T.V.

PICTORIAL PRESENTATION IS GIVEN

DRAW THREE VIEWS OF THIS OBJECT

BY FIRST ANGLE PROJECTION METHOD

ORTHOGRAPHIC PROJECTIONS

1

O


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FOR T.V.


X Y

FRONT VIEW

TOP VIEW

L.H.SIDE VIEW

3





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ZSTUDY

ILLUSTRATIONS

X Y

50

20

25

25 20

FOR T.V.




8


FRONT VIEW

TOP VIEW


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FOR T.V.


DRAW THREE VIEWS OF THIS OBJECTBY FIRST ANGLE PROJECTION METHOD

10ORTHOGRAPHIC PROJECTIONS

FRONT VIEW

TOP VIEW

L.H.SIDE VIEW

X Y


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ZSTUDYILLUSTRATIONS

SV

TV

yx

FV

30

30

10

30 10 30

ALL VIEWS IDENTICAL

FOR T.V.




14


FOR T.V.


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DRAW THREE VIEWS OF THIS OBJECTBY FIRST ANGLE PROJECTION METHOD

16ORTHOGRAPHIC PROJECTIONS

x y

FV SV

ALL VIEWS IDENTICAL

40 60

60

40

10

TOP VIEW


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40 20

30 SQUARE

20

50

60

30

10

F.V.S.V.

O


DRAW FV AND SV OF THIS OBJECTBY FIRST ANGLE PROJECTION METHOD

17


FRONT VIEW L.H.SIDE VIEW

X Y



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TO DRAW PROJECTIONS OF ANY OBJECT,

ONE MUST HAVE FOLLOWING INFORMATIONA)OBJECT

{ WITH ITS DESCRIPTION, WELL DEFINED.}

B) OBSERVER{ ALWAYS OBSERVING PERPENDICULAR TO RESP. REF.PLANE}.

C) LOCATION OF OBJECT,{ MEANS ITS POSITION WITH REFFERENCE TO H.P. & V.P.}

TERMS ABOVE & BELOW WITH RESPECTIVE TO H.P.AND TERMS INFRONT & BEHIND WITH RESPECTIVE TO V.P

FORM 4 QUADRANTS.OBJECTS CAN BE PLACED IN ANY ONE OF THESE 4 QUADRANTS.

IT IS INTERESTING TO LEARN THE EFFECT ON THE POSITIONS OF VIEWS ( FV, TV )OF THE OBJECT WITH RESP. TO X-Y LINE, WHEN PLACED IN DIFFERENT QUADRANTS.

ORTHOGRAPHIC PROJECTIONSOF POINTS, LINES, PLANES, AND SOLIDS.

STUDY ILLUSTRATIONS GIVEN ON HEXT PAGES AND NOTE THE RESULTS.TO MAKE IT EASYHERE A POINT A IS TAKEN AS AN OBJECT. BECAUSE ITS ALL VIEWS ARE JUST POINTS.


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NOTATIONS

FOLLOWING NOTATIONS SHOULD BE FOLLOWED WHILE NAMEING

DIFFERENT VIEWS IN ORTHOGRAPHIC PROJECTIONS.

ITS FRONT VIEW a a b

SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWEDINCASE NUMBERS, LIKE 1, 2, 3 ARE USED.

OBJECT POINT A LINE AB

ITS TOP VIEW a a b

ITS SIDE VIEW a a b


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X

Y

1ST Quad.2nd Quad.

3rd Quad. 4th Quad.

X Y

VP

HP

Observer

THIS QUADRANT PATTERN,IF OBSERVED ALONG X-Y LINE ( IN RED ARROW DIRECTION)WILL EXACTLY APPEAR AS SHOWN ON RIGHT SIDE AND HENCE,

IT IS FURTHER USED TO UNDERSTAND ILLUSTRATION PROPERLLY.

VP POINT A INST QUADRANT

POINT A INPoint A is


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HP

VP

a

a

A

1ST QUADRANT

OBSERVER

VP

HP

2ND QUADRANT

OBSERVER

a

a

A

OBSERVER

a

a

POINT A IN3RD QUADRANT

HP

VP

A

OBSERVER

a

aPOINT A IN4TH QUADRANT

HP

VP

A

Placed Indifferentquadrants

and its Fv & Tv

are brought insame plane for

Observer to seeclearly.

Fv is visible asit is a view on

VP. But as Tv isis a view on Hp,

it is rotateddownward 900,

In clockwisedirection.TheIn front part of

Hp comes belowxy line and thepart behind Vpcomes above.

Observe andnote theprocess.

PROJECTIONS OF A POINT IN FIRST QUADRANT.


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A

a

aA

a

a

Aa

a

X

Y

X

Y

X

Y

For TvFor Tv

For Tv

POINT A ABOVE HP& INFRONT OF VP

POINT A IN HP& INFRONT OF VP

POINT A ABOVE HP& IN VP

PROJECTIONS OF A POINT IN FIRST QUADRANT.

PICTORIALPRESENTATION

PICTORIALPRESENTATION

ORTHOGRAPHIC PRESENTATIONSOF ALL ABOVE CASES.

X Y

a

a

VP

HP

X Y

a

VP

HP

a X Y

a

VP

HP

a

Fv above xy,Tv below xy.

Fv above xy,Tv on xy.

Fv on xy,Tv below xy.

PROJECTIONS OF STRAIGHT LINES


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SIMPLE CASES OF THE LINE

1. A VERTICAL LINE ( LINE PERPENDICULAR TO HP & // TO VP)

2. LINE PARALLEL TO BOTH HP & VP.

3. LINE INCLINED TO HP & PARALLEL TO VP.

4. LINE INCLINED TO VP & PARALLEL TO HP.

5. LINE INCLINED TO BOTH HP & VP.

STUDY ILLUSTRATIONS GIVEN ON NEXT PAGE

SHOWING CLEARLY THE NATURE OF FV & TV

OF LINES LISTED ABOVE AND NOTE RESULTS.

PROJECTIONS OF STRAIGHT LINES.

INFORMATION REGARDING A LINE meansITS LENGTH,

POSITION OF ITS ENDS WITH HP & VP

ITS INCLINATIONS WITH HP & VP WILL BE GIVEN.AIM:- TO DRAW ITS PROJECTIONS - MEANS FV & TV.

For Tv Orthographic Pattern(Pi t i l P t ti )


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X

Y

X

Y

b

a

b

a

a b

a

b

B

A

TV

FV

A

B

X Y

H.P.

V.P.a

b

a b

Fv

Tv

X Y

H.P.

V.P.

a b

a bFv

Tv

For Tv

Note:Fv is a vertical line

Showing True Length

&

Tv is a point.

Note:Fv & Tv both are

// to xy

&

both show T. L.

1.

2.

A Line

perpendicular

to Hp

&

// to Vp

A Line

// to Hp

&

// to Vp

Orthographic Pattern

(Pictorial Presentation)

(Pictorial Presentation)


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A Line inclined to Hp

andparallel to Vp

(Pictorial presentation)X

Y

A

B

b

a

b

a

A Line inclined to Vpand

parallel to Hp

(Pictorial presentation)

a b

a

b

BA

X Y

H.P.

V.P.

T.V.a b

a

b

X Y

H.P.

V.P.

a

b

a b

Tv

Fv

Tv inclined to xy

Fv parallel to xy.

3.

4.

Fv inclined to xy

Tv parallel to xy.

Orthographic Projections

For Tv For Tv


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X

Y

a

b

a b

B

A

T.V.

X

Ya

b

a b

T.V.

B

A

X Y

H.P.

V.P.

a

b

FV

TV

a

b

A Line inclined to bothHp and Vp

(Pictorial presentation)

5.

Note These Facts:-Both Fv & Tv are inclined to xy.

(No view is parallel to xy)

Both Fv & Tv are reduced lengths.(No view shows True Length)

Orthographic ProjectionsFv is seen on Vp clearly.

To see Tv clearly, HP isrotated 900downwards,

Hence it comes below xy.

On removal of objecti.e. Line AB

Fv as a image on Vp.

Tv as a image on Hp,

Note the procedure Note the procedureOrthographic Projections


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X Y

H.P.

V.P.

X Y

H.P.

V.P.

a

b

TV

a

b

FV

TV

b2

b1

TL

X Y

H.P.

V.P.

a

b

FV

TV

a

b

Here TV (ab) is not // to XY lineHence its corresponding FV

a b is notshowingTrue Length &

True Inclination with Hp.

In this sketch, TV is rotatedand made // to XY line.

Hence its corresponding

FV a b1Is showingTrue Length

&

True Inclination with Hp.

pWhen Fv & Tv known,

How to find True Length.(Views are rotated to determineTrue Length & its inclinations

with Hp & Vp).

pWhen True Length is known,

How to locate Fv & Tv.(Component a-1of TL is drawn

which is further rotatedto determine Fv)

1a

a

b

1

b

b1

b1

Orthographic ProjectionsMeans Fv & Tv of Line AB

are shown below,

with their apparent Inclinations

&

Herea -1 is componentof TL ab1 gives length ofFv.

Hence it is brought Up toLocus of a and further rotatedto get pointb. a bwill be Fv.

Similarly drawing component

of other TL(a b1) Tv can be drawn.

The most important diagram showing graphical relations

ll i t t t f thi t i Important1) True Length ( TL)a b1 & a b

2) A l f TL ith H


48/159

among all important parameters of this topic.

Study and memorize it as a CIRCUIT DIAGRAMAnd use in solving various problems.

True Length is never rotated. Its horizontal component

is drawn & it is further rotated to locate view.

Views are always rotated, made horizontal & furtherextended to locate TL, &

Also Remember

ImportantTEN parameters

to be rememberedwith Notations

used here onward

2) Angle of TL with Hp -3) Angle of TL with Vp

4) Angle of FV with xy

5) Angle of TV with xy

6) LTV (length of FV) Component (a-1)

7) LFV (length of TV) Component (a-1)

8) Position of A- Distances of a & a from xy

9) Position of B- Distances of b & b from xy

10) Distance between End Projectors

X Y

H.P.

V.P.

1a

b

b1

LFV

a

b

1

b1

LTV

Distance between

End Projectors.

& Construct with a & Construct with a

b & b1 on same locus.

b & b1 on same locus.

NOTE this

GROUP (A)GENERAL CASES OF THE LINE INCLINED TO BOTH HP & VP


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a

b

a

b

X Y

b1

b1

GENERAL CASES OF THE LINE INCLINED TO BOTH HP & VP( based on 10 parameters).PROBLEM 1)

Line AB is 75 mm long and it is 300 &

400 Inclined to Hp & Vp respectively.

End A is 12mm above Hp and 10 mm

in front of Vp.

Draw projections. Line is in 1st quadrant.

SOLUTION STEPS:1) Draw xy line and one projector.2) Locate a 12mm above xy line

& a 10mm below xy line.3) Take 300angle from a & 400 from

a and mark TL I.e. 75mm on bothlines. Name those points b1 and b1respectively.

4) Join both points with a and a resp.5) Draw horizontal lines (Locus) from

both points.6) Draw horizontal component of TL

a b1 from point b1 and name it 1.( the length a-1 gives length of Fvas we have seen already.)

7) Extend it up to locus of a and

rotating a as center locate b as

shown. Join a b as Fv.8) From b drop a projector down

ward & get point b. Join a & b

I.e. Tv.

1LFV

TL

TL

FV

TV

PROBLEM 2:


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X y

a

a

b1

1

b1b

LFV

550

b

PROBLEM 2:

Line AB 75mm long makes 450inclination with Vp while its Fv makes 550.

End A is 10 mm above Hp and 15 mm in front of Vp.If line is in 1st quadrant

draw its projections and find its inclination with Hp.

LOCUS OF b

LOCUS OF

Solution Steps:-1.Draw x-y line.2.Draw one projector for a & a3.Locate a10mm above x-y &Tv a15 mm below xy.

4.Draw a line 450 inclined to xyfrom point aand cut TL 75 mmon it and name that point b1Draw locus from point b1

5.Take 550

angle from afor Fvabove xy line.6.Draw a vertical line from b1

up to locus of a and name it 1.It is horizontal component ofTL & is LFV.

7.Continue it to locus of aandrotate upward up to the lineof Fv and name it b.This a b

line is Fv.8. Drop a projector from b on

locus from point b1 andname intersecting point b.Line a bis Tv of line AB.

9.Draw locus from band fromawith TL distance cut point b1

10.Join a b1as TL and measureits angle at a.

It will be true angle of line with HP.

PROBLEM 3:

Fv of line AB is 500 inclined to xy and measures 55


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Xa

y

a

b

500

b

600

b1

b1

Fv of line AB is 500 inclined to xy and measures 55

mm long while its Tv is 600 inclined to xy line. If

end A is 10 mm above Hp and 15 mm in front of

Vp, draw its projections,find TL, inclinations of line

with Hp & Vp.

SOLUTION STEPS:1.Draw xy line and one projector.

2.Locate a 10 mm above xy and

a 15 mm below xy line.

3.Draw locus from these points.

4.Draw Fv 500to xy from a and

mark b Cutting 55mm on it.

5.Similarly draw Tv 600 to xy

from a & drawing projector from b

Locate point b and join a b.

6.Then rotating views as shown,

locate True Lengths ab1 & ab1

and their angles with Hp and Vp.


52/159

X Ya

1

a

b1

LTV

b1

1

b

b

LFV

PROBLEM 4 :-Line AB is 75 mm long .Its Fv and Tv measure 50 mm & 60 mm long respectively.End A is 10 mm above Hp and 15 mm in front of Vp. Draw projections of line ABif end B is in first quadrant.Find angle with Hp and Vp.


2.Locate a 10 mm above xy and

a 15 mm below xy line.


4.Cut 60mm distance on locus of a

& mark 1 on it as it is LTV.

5.Similarly Similarly cut 50mm on

locus of a and mark point 1 as it is LFV.

6.From 1 draw a vertical line upward

and from a taking TL ( 75mm ) in

compass, mark b1 point on it.

Join a b1 points.

7. Draw locus from b1

8. With same steps below get b1 pointand draw also locus from it.

9. Now rotating one of the components

I.e. a-1 locate b and join a with it

to get Fv.

10. Locate tv similarly and measure

Angles &

PROBLEM 5 :-

T V of a 75 mm long Line CD measures 50 mm


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X Yc

c

LOCUS OFd & d1d d1

d d1

LOCUS OFd & d1

T.V. of a 75 mm long Line CD, measures 50 mm.

End C is in Hp and 50 mm in front of Vp.

End D is 15 mm in front of Vp and it is above Hp.

Draw projections of CD and find angles with Hp and Vp.


2.Locate c on xy and

c 50mm below xy line.


4.Draw locus ofd 15 mm below xy

5.Cut 50mm & 75 mm distances onlocus ofd from c and mark points

d & d1 as these are Tv and line CD

lengths resp.& join both with c.

6.From d1 draw a vertical line upward

up to xy I.e. up to locus ofc and

draw an arc as shown.

7 Then draw one projector from d to

meet this arc in d point & join c d8. Draw locus ofd and cut 75 mm

on it from c as TL

9.Measure Angles &

GROUP (B)


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TRACES OF THE LINE:-

THESE ARE THE POINTS OF INTERSECTIONS OF A LINE ( OR ITS EXTENSION )

WITH RESPECTIVE REFFERENCE PLANES.

A LINE ITSELF OR ITS EXTENSION, WHERE EVER TOUCHES H.P.,

THAT POINT IS CALLED TRACE OF THE LINE ON H.P.( IT IS CALLED H.T.)

SIMILARLY, A LINE ITSELF OR ITS EXTENSION, WHERE EVER TOUCHES V.P.,

THAT POINT IS CALLED TRACE OF THE LINE ON V.P.( IT IS CALLED V.T.)

V.T.:- It is a point on Vp.

Hence it is called Fvof a point in Vp.Hence its Tvcomes on XY line.( Here onward named as v )

H.T.:- It is a point on Hp.Hence it is called Tvof a point in Hp.

Hence its Fvcomes on XY line.( Here onward named as h )

GROUP (B)PROBLEMS INVOLVING TRACES OF THE LINE.


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1. Begin with FV. Extend FV up to XY line.

2. Name this point h( as it is a Fv of a point in Hp)

3. Draw one projector fromh.

4. Now extend Tv to meet this projector.

This point is HT

STEPS TO LOCATE HT.

(WHEN PROJECTIONS ARE GIVEN.)

1. Begin with TV. Extend TV up to XY line.

2. Name this point v

( as it is a Tv of a point in Vp)3. Draw one projector from v.

4. Now extend Fv to meet this projector.

This point is VT

STEPS TO LOCATE VT.

(WHEN PROJECTIONS ARE GIVEN.)

h

HTVT

v

a

x y

a

bObserve & note :-1. Points h & v always on x-y line.

2. VT & v always on one projector.

3. HT & h always on one projector.

4. FV - h- VT always co-linear.

5. TV - v - HT always co-linear.

These points are used to

solve nextthree problems.

PROBLEM 6 :- Fv of line AB makes 450 angle with XY line and measures 60 mm.i k 0 i h li d i b d i i


56/159

x y

b b1

a

v

VT

a

b

h

b1

300

450

Lines Tv makes 300with XY line. End A is 15 mm above Hp and its VT is 10 mm

below Hp. Draw projections of line AB,determine inclinations with Hp & Vp and locate HT, VT.

15

10

SOLUTION STEPS:-Draw xy line, one projector and

locate fv a 15 mm above xy.

Take 450 angle from a and

marking 60 mm on it locate point b.

Draw locus ofVT, 10 mm below xy

& extending Fv to this locus locate VT.

as fv-h-vt lie on one st.line.

Draw projector from vt, locate v on xy.From v take 300 angle downward as

Tv and its inclination can begin with v.

Draw projector from b and locate b I.e.Tv point.

Now rotating views as usual TL and

its inclinations can be found.

Name extension of Fv, touching xy as h

and below it, on extension of Tv, locate HT.

PROBLEM 7 :


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a

b

3045

10

LOCUS OF b & b1

X Y

450

VT

v

HT

h

LOCUS OF b & b1

100

a

b

b1

b1

PROBLEM 7 :One end of line AB is 10mm above Hp and other end is 100 mm in-front of Vp.Its Fv is 450inclined to xy while its HT & VT are 45mm and 30 mm below xy respectively.Draw projections and find TL with its inclinations with Hp & VP.

SOLUTION STEPS:-Draw xy line, one projector and

locate a 10 mm above xy.

Draw locus 100 mm below xy for points b & b1Draw loci for VT and HT, 30 mm & 45 mm

below xy respectively.

Take 450 angle from a and extend that line backward

to locate h and VT, & Locate v on xy above VT.

Locate HT below h as shown.

Then join v HT and extend to get top view end b.

Draw projector upward and locate b Make a b & ab dark.

Now as usual rotating views find TL and its inclinations.

PROBLEM 8 :-Projectors drawn from HT and VT of a line AB80 t d th d f it d 50 t


58/159

X y

HT

VT

h

a

v

b

a

b

80

50

b1

b 1

10

35

55

Locus of a

are 80 mm apart and those drawn from its ends are 50 mm apart.

End A is 10 mm above Hp, VT is 35 mm below Hp

while its HT is 45 mm in front of Vp. Draw projections,

locate traces and find TL of line & inclinations with Hp and Vp.

SOLUTION STEPS:-1.Draw xy line and two projectors,

80 mm apart and locate HT & VT ,

35 mm below xy and 55 mm above xy

respectively on these projectors.2.Locate h and v on xy as usual.

3.Now just like previous two problems,

Extending certain lines complete Fv & Tv

And as usual find TL and its inclinations.

I t d f id i & j ti f fi t i t


59/159

b1

a

VT

vX Y

b

a

b

b1

Then from point v & HTangles can be drawn.

&From point VT & hangles can be drawn.&

&

Instead of considering a & a as projections of first point,if v & VT are considered as first point , then true inclinations of line with

Hp & Vp i.e. angles & can be constructed with points VT & V respectively.

THIS CONCEPT IS USED TO SOLVENEXT THREE PROBLEMS.

PROBLEM 9 :-


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Line AB 100 mm long is 300 and 450 inclined to Hp & Vp respectively.

End A is 10 mm above Hp and its VT is 20 mm below Hp

.Draw projections of the line and its HT.

X Y

VT

v10

20

Locus of a & a1

(300)

(450)

a1

b1

b1

a1

b

a

b

a

FV

TV

HT

hSOLUTION STEPS:-Draw xy, one projector

and locate on it VT and V.

Draw locus ofa 10 mm above xy.

Take 300 from VT and draw a line.Where it intersects with locus ofa

name it a1 as it is TL of that part.

From a1 cut 100 mm (TL) on it and locate point b1

Now from v take 450 and draw a line downwards

& Mark on it distance VT-a1 I.e.TL of extension & name it a1

Extend this line by 100 mm and mark point b1.Draw its component on locus ofVT

& further rotate to get other end of Fv i.e.b

Join it with VT and mark intersection point

(with locus ofa1 ) and name it a

Now as usual locate points a and b andh and HT.

PROBLEM 10 :-


61/159

A line AB is 75 mm long. Its Fv & Tv make 450 and 600 inclinations with X-Y line resp

End A is 15 mm above Hp and VT is 20 mm below Xy line. Line is in first quadrant.

Draw projections, find inclinations with Hp & Vp. Also locate HT.

X Y

VT

v15

20

Locus of a & a1 a1

b1

b1

a1

b

a

b

a

FV

TV

HT

h

450

600

SOLUTION STEPS:-Similar to the previous only change

is instead of lines inclinations,

views inclinations are given.

So first take those angles from VT & v

Properly, construct Fv & Tv of extension,then determine its TL( V-a1)

and on its extension mark TL of line

and proceed and complete it.


62/159

PROBLEM 11 :- The projectors drawn from VT & end A of line AB are 40mm apart.End A is 15mm above Hp and 25 mm in front of Vp. VT of line is 20 mm below Hp.

If line is 75mm long, draw its projections, find inclinations with HP & Vp

X Y

40mm

15

2025

v

VT

a

a

a1

b1b

b b1

Draw two projectors for VT & end A

Locate these points and then

YES !YOU CAN COMPLETE IT.

GROUP (C)CASES OF THE LINES IN A V P A I P & PROFILE PLANE


63/159

CASES OF THE LINES IN A.V.P., A.I.P. & PROFILE PLANE.

a

b Line AB is in AIP as shown in above figure no 1.

Its FV (ab) is shown projected on Vp.(Looking in arrow direction)Here one can clearly see that the

Inclination of AIP with HP = Inclination of FV with XY line

Line AB is in AVP as shown in above figure no 2..

Its TV (a b) is shown projected on Hp.(Looking in arrow direction)

Here one can clearly see that the

Inclination of AVP with VP = Inclination of TV with XY line

A.V.P.

A

B

a b

LINE IN A PROFILE PLANE( MEANS IN A PLANE PERPENDICULAR TO BOTH HP & VP)


64/159

PPVP

HP

a

b

a

b

a

b

X Y

FV

TV

LSV

A

B

a

b

a

b

For T.V.

Results:-1. TV & FV both are vertical, hence arrive on one single projector.

2. Its Side View shows True Length ( TL)

3. Sum of its inclinations with HP & VP equals to 900 (

4. Its HT & VT arrive on same projector and can be easily located

From Side View.

+ = 900 )

OBSERVE CAREFULLY ABOVE GIVEN ILLUSTRATION AND 2ndSOLVED PROBLEM.

ORTHOGRAPHIC PATTERN OF LINE IN PROFILE PLANE

HT

VT


65/159

PROBLEM 12 :- Line AB 80 mm long, makes 300 angle with Hpand lies in an Aux.Vertical Plane 450 inclined to Vp.

End A is 15 mm above Hp and VT is 10 mm below X-y line.

Draw projections, fine angle with Vp and Ht.

VT

vX Y

a

b

a

b

a1

b1

Locus of b

Locus of b

10

15

HT

h

b1

AVP 450 to VP

450

Locus of a & a1

Simply consider inclination of AVP

as inclination of TV of our line,

well then?

You sure can complete it

as previous problems!

Go ahead!!


66/159

PROBLEM 13 :- A line AB, 75mm long, has one end A in Vp. Other end B is 15 mm above Hpand 50 mm in front of Vp.Draw the projections of the line when sum of its

Inclinations with HP & Vp is 900

, means it is lying in a profile plane.Find true angles with ref.planes and its traces.

a

b

HT

VT

X Y

a

b

Side View

( True Length )

a

b

(HT)

(VT)

HP

V

PFront view

top view

SOLUTION STEPS:-

After drawing xy line and one projector

Locate top view of A I.e point a on xy asIt is in Vp,

Locate Fv of B i.e.b15 mm above xy as

it is above Hp.and Tv of B i.e. b, 50 mm

below xy asit is 50 mm in front of Vp

Draw side view structure of Vp and Hp

and locate S.V. of point B i.e. b

From this point cut 75 mm distance on Vp andMark a as A is in Vp. (This is also VT of line.)

From this point draw locus to left & get a

Extend SV up to Hp. It will be HT. As it is a Tv

Rotate it and bring it on projector of b.

Now as discussed earlier SV gives TL of line

and at the same time on extension up to Hp & Vp

gives inclinations with those panes.

APPLICATIONS OF PRINCIPLES OF PROJECTIONS OF LINES


67/159

APPLICATIONS OF PRINCIPLES OF PROJECTIONS OF LINESIN SOLVING CASES OF DIFFERENT PRACTICAL SITUATIONS.

In these types of problems some situation in the field

orsome object will be described .Its relation with Ground ( HP )

Anda Wall or some vertical object ( VP ) will be given.

Indirectly information regarding Fv & Tv of some line or lines,inclined to both reference Planes will be given

andyou are supposed to draw its projections

andfurther to determine its true Length and its inclinations with ground.

Here various problems along withactual pictures of those situations are given

for you to understand those clearly.Now looking for views in given ARROW directions,

YOU are supposed to draw projections & find answers,Off course you must visualize the situation properly.

CHECK YOUR ANSWERSWITH THE SOLUTIONS

GIVEN IN THE END.

ALL THE BEST !!

PROBLEM NO.24T.V. of a 75 mm long Line CD, measures 50 mm.


68/159

X Y

c

c

LOCUS OFd & d1d d1

d d1

LOCUS OFd & d1

End C is 15 mm below Hp and 50 mm in front of Vp.

End D is 15 mm in front of Vp and it is above Hp.

Draw projections of CD and find angles with Hp and Vp.

SOME CASES OF THE LINEIN DIFFERENT QUADRANTS.

REMEMBER:BELOW HP- Means- Fv below xy

BEHIND V p- Means- Tv above xy.

PROJECTIONS OF PLANES


69/159

In this topic various plane figures are the objects.

What will be given in the problem?

1. Description of the plane figure.2. Its position with HP and VP.

In which manner its position with HP & VP will be described?

1.Inclination of its SURFACEwith one of the reference planes will be given.2.Inclination of one of itsEDGESwithotherreference plane will be given

(Hence this will be a case of an object inclined to both reference Planes.)

To draw their projections means F.V, T.V. & S.V.

What is usually asked in the problem?

Study the illustration showingsurface & side inclination given on next page.

CASE OF A RECTANGLE OBSERVE AND NOTE ALL STEPS.

SURFACE PARALLEL TO HP SURFACE INCLINED TO HP ONE SMALL SIDE INCLINED TO VP


70/159

HP

VPVPVP

a dcb

HP

a

b c

d

a1

d1c

1

b1

HP

a1

b1 c1

d1

SURFACE PARALLELTO HPPICTORIAL PRESENTATION

SURFACE INCLINEDTO HPPICTORIAL PRESENTATION

ONE SMALL SIDE INCLINED TO VPPICTORIAL PRESENTATION

ORTHOGRAPHICTV-True Shape

FV- Line // to xy

ORTHOGRAPHICFV- Inclined to XY

TV- Reduced Shape

ORTHOGRAPHICFV- Apparent Shape

TV-Previous Shape

A B C

PROCEDURE OF SOLVING THE PROBLEM:IN THREE STEPS EACH PROBLEM CAN BE SOLVED:( As Shown In Previous Illustration )


71/159

IN THREE STEPS EACH PROBLEM CAN BE SOLVED:( As Shown In Previous Illustration )STEP 1.Assume suitable conditions & draw Fv & Tv of initial position.

STEP 2.Now consider surface inclination & draw 2nd Fv & Tv.

STEP 3.After this,consider side/edge inclination and draw 3rd ( final) Fv & Tv.

ASSUMPTIONS FOR INITIAL POSITION:(Initial Position means assuming surface // to HP or VP)

1.If in problem surface is inclined to HP assume it // HP

Or If surface is inclined to VP assume it // to VP

2. Now if surface is assumed // to HP- Its TV will show True Shape.

And If surface is assumed // to VPIts FV will show True Shape.3. Hence begin with drawing TV or FV as True Shape.

4. While drawing this True Shape

keep one side/edge ( which is making inclination) perpendicular to xy line

( similar to pair no. on previous page illustration ).A

B

Now Complete STEP 2. By making surface inclined to the resp plane & project its other view.(Ref. 2ndpair on previous page illustration)

C

Now Complete STEP 3. By making side inclined to the resp plane & project its other view.(Ref. 3ndpair on previous page illustration)

APPLY SAME STEPS TO SOLVE NEXT PROBLEMS

Problem 1: Read problem and answer following questions


72/159

X Y

a

b c

d

ab

cd

a1

b1 c1

d1

ab

dc c1 d1

b1 a1450

300

Rectangle 30mm and 50mm

sides is resting on HP on one

small side which is 300 inclined

to VP,while the surface of the

plane makes 450 inclination with

HP. Draw its projections.

p g q

1. Surface inclined to which plane? ------- HP

2. Assumption for initial position? ------// to HP

3. So which view will show True shape? --- TV

4. Which side will be vertical? ---One small side.

Hence begin with TV, draw rectangle below X-Y

drawing one small side vertical.

Surface // to Hp Surface inclined to Hp

Side

Inclinedto Vp

Problem 2:A 300 600 set square of longest side

Read problem and answer following questions

1 S f i li d t hi h l ? VP


73/159

A 300 600 set square of longest side

100 mm long, is in VP and 300 inclined

to HP while its surface is 450 inclined

to VP.Draw its projections

(Surface & Side inclinations directly given)

1 .Surface inclined to which plane? ------- VP

2. Assumption for initial position? ------// to VP

3. So which view will show True shape? --- FV

4. Which side will be vertical? ------longest side.

c1

X Y300

450

a1

b1

c1

a

c

a

ab1

b

b

a1b

c

a1

b1

c1

c

Hence begin with FV, draw triangle above X-Y

keeping longest side vertical.

Surface // to Vp Surface inclined to Vp

side inclined to Hp

Problem 3:A 300 600 set square of longest side

Read problem and answer following questions1 .Surface inclined to which plane? ------- VP

2 A ti f i iti l iti ? // t VP


74/159

cc1

X Y450

a1

b1

c1

a

c

a

ab1

b

b

a1b

a1

b1

c1

c

35

10

100 mm long is in VP and its surface

450 inclined to VP. One end of longest

side is 10 mm and other end is 35 mm

above HP. Draw its projections

(Surface inclination directly given.Side inclination indirectly given)

2. Assumption for initial position? ------// to VP

3. So which view will show True shape? --- FV

4. Which side will be vertical? ------longest side.

Hence begin with FV, draw triangle above X-Ykeeping longest side vertical.

First TWO steps are similar to previous problem.Note the manner in which side inclination is given.End A 35 mm above Hp & End B is 10 mm above Hp.So redraw 2nd Fv as final Fv placing these ends as said.

Read problem and answer following questions1. Surface inclined to which plane? ------- HP

Problem 4:A regular pentagon of 30 mm sides is


75/159

p



4. Which side will be vertical? --------any side.

Hence begin with TV,draw pentagon below

X-Y line, taking one side vertical.

g p g

resting on HP on one of its sides with its

surface 450 inclined to HP.

Draw its projections when the side in HP

makes 300 angle with VP

ab d

b1

d

c1

a

ce

b

c

d1

b1

a1

e1 c1

d1

a1

b1

c1

d1

d

ab

ce

e1

e1

a1X Y450

300e

SURFACE AND SIDE INCLINATIONSARE DIRECTLY GIVEN.

Problem 5:A l t f 30 id i ti


1 Surface inclined to which plane? HP


76/159

A regular pentagon of 30 mm sides is resting

on HP on one of its sides while its opposite

vertex (corner) is 30 mm above HP.

Draw projections when side in HP is 300

inclined to VP.




4. Which side will be vertical? --------any side.

Hence begin with TV,draw pentagon below

X-Y line, taking one side vertical.

b

d

a

ce

a1

b1

c1d1

e1

b1

c1

d1

a1

e1

b1

e1 c1

d1

a1X Yab dce

30

a

b

c

d

e

300

SURFACE INCLINATION INDIRECTLY GIVENSIDE INCLINATION DIRECTLY GIVEN:

ONLY CHANGE is

the manner in which surface inclination is described:

One side on Hp & its opposite corner 30 mm above Hp.

Hence redraw 1st Fv as a 2nd Fv making above arrangement.

Keep ab on xy & d 30 mm above xy.

a bd c

Y

b1c1

d

Problem 8: A circle of 50 mm diameter isresting on Hp on end A of its diameter AC


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a

d

c

b

X Y

a1

b1

d1

c1

450300 a1

d 1

a1

b1

d1

c1a

d

c

b

a bd c

300

a1

b1c1

d1

resting on Hp on end A of its diameter AC

which is 300inclined to Hp while its Tv

is 450inclined to Vp.Draw its projections.

Problem 9: A circle of 50 mm diameter isresting on Hp on end A of its diameter AC

which is 300 inclined to Hp while it makes

450inclined to Vp. Draw its projections.





4. Which diameter horizontal? ---------- AC

Hence begin with TV,draw rhombus below

X-Y line, taking longer diagonal // to X-Y

The difference in these two problems is in step 3 only.In problem no.8 inclination of Tv of that AC isgiven,It could be drawn directly as shown in 3rd step.

While in no.9 angle of AC itself i.e. its TL, isgiven. Hence here angle of TL is taken,locus of c1Is drawn and then LTV I.e. a1 c1 is marked andfinal TV was completed.Study illustration carefully.

Note the difference inconstruction of 3rd step

in both solutions.

Problem 10: End A of diameter AB of a circle is in HP




78/159

Problem 10: End A of diameter AB of a circle is in HPA nd end B is in VP.Diameter AB, 50 mm long is

300 & 600 inclined to HP & VP respectively.

Draw projections of circle.

The problem is similar to previous problem of circle no.9.But in the 3rd step there is one more change.Like 9th problem True Length inclination of dia.AB is definitely expectedbut if you carefully note - the the SUM of its inclinations with HP & VP is 900.Means Line AB lies in a Profile Plane.

Hence its both Tv & Fv must arrive on one single projector.So do the construction accordingly AND note the case carefully..

SOLVE SEPARATELYON DRAWING SHEET

GIVING NAMES TO VARIOUS

POINTS AS USUAL,

AS THE CASE IS IMPORTANT

X Y300

600

p



4. Which diameter horizontal? ---------- AB

Hence begin with TV,draw CIRCLE below

X-Y line, taking DIA. AB // to X-Y

Problem 11:A hexagonal lamina has its one side in HP and


1 Surface inclined to which plane? HP


79/159

As 3rd

stepredraw 2nd Tv keepingside DE on xy line.Because it is in VPas said in problem.

X Y

a

b

c

d

e

f

e ago a a a as s o e s de a d

Its apposite parallel side is 25mm above Hp and

In Vp. Draw its projections.

Take side of hexagon 30 mm long.

ONLY CHANGE is the manner in which surface inclination

is described:

One side on Hp & its opposite side 25 mm above Hp.

Hence redraw 1st Fv as a 2nd Fv making above arrangement.

Keep ab on xy & de 25 mm above xy.

25

f edcba

a1

b1

c1

d1

e1

f1

c1

b1a1

f1

d1e1

f1

a1

c1

b1

d1e1




4. Which diameter horizontal? ---------- AC

Hence begin with TV,draw rhombus below

X-Y line, taking longer diagonal // to X-Y

FREELY SUSPENDED CASES.

IMPORTANT POINTS


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A B

C

H

H/3

G

X Y

a

b

c

g

b a,g c 450

a1

c1

b1g1

1.In this case the plane of the figure always remainsperpendicular to Hp.

2.It may remain parallel or inclined to Vp.

3.Hence TVin this case will be always aLINE view.

4.Assuming surface // to Vp, draw true shape in suspended position as FV.

(Here keep line joiningpoint of contact & centroidof fig. vertical)5.Always begin with FV as a True Shape but in a suspended position.

AS shown in 1st FV.

IMPORTANT POINTS

Problem 12:An isosceles triangle of 40 mm longbase side, 60 mm long altitude Is

freely suspended from one corner ofBase side.Its plane is 450 inclined toVp. Draw its projections.

Similarly solve next problemof Semi-circle

First draw a given triangleWith given dimensions,

Locate its centroid positionAnd

join it with point of suspension.

Problem 13:A semicircle of 100 mm diameter

1.In this case the plane of the figure always remainsperpendicular to Hp.

2 It may remain parallel or inclined to Vp

IMPORTANT POINTS


81/159

G

A

P

20 mm

CG

X Y

e

c

d

b

a

p

g

b c a p,g d e

:A semicircle of 100 mm diameteris suspended from a point on its

straight edge 30 mm from the midpoint

of that edge so that the surface makes

an angle of 450 with VP.

Draw its projections.

First draw a given semicircle

With given diameter,Locate its centroid position

And

join it with point of suspension.

2.It may remain parallel or inclined to Vp.

3.Hence TVin this case will be always aLINE view.

4.Assuming surface // to Vp, draw true shape in suspended position as FV.

(Here keep line joiningpoint of contact & centroidof fig. vertical)5.Always begin with FV as a True Shape but in a suspended position.

AS shown in 1st FV.

To determine true shape of plane figure when its projections are given.BY USING AUXILIARY PLANE METHOD


82/159

BY USING AUXILIARY PLANE METHOD

WHAT WILL BE THE PROBLEM?

Description of final Fv & Tv will be given.

You are supposed to determine true shape of that plane figure.

Follow the below given steps:1. Draw the given Fv & Tv as per the given information in problem.

2. Then among all lines of Fv & Tv select a line showing True Length (T.L.)

(Its other view must be // to xy)

3. Draw x1-y1 perpendicular to this line showing T.L.

4. Project view on x1-y1 ( it must be a line view)

5. Draw x2-y2 // to this line view & project new view on it.It will be the required answer i.e. True Shape.

The facts you must know:-

If you carefully study and observe the solutions of all previous problems,

You will find

IF ONE VIEW IS A LINE VIEW & THAT TOO PARALLEL TO XY LINE,

THEN AND THEN ITS OTHER VIEW WILL SHOW TRUE SHAPE:

NOW FINAL VIEWS ARE ALWAYS SOME SHAPE, NOT LINE VIEWS:

SO APPLYING ABOVE METHOD:

WE FIRST CONVERT ONE VIEW IN INCLINED LINE VIEW .(By using x1y1 aux.plane)

THEN BY MAKING IT // TO X2-Y2 WE GET TRUE SHAPE.

Study Next

Four Cases

Problem 14 Tv is a triangle abc. Ab is 50 mm long, angle cab is 300 and angle cba is 650.abc is a Fv. a is 25 mm, b is 40 mm and c is 10 mm above Hp respectively. Draw projections


83/159

X Y

a

c

b

C

b

a

10

15

15

X1

Y1

C1

b1a1

a1

b1

c1

X2

Y2

p p y p j

of that figure and find its true shape.

300 650

50 mm

As per the procedure-1.First draw Fv & Tv as per the data.2.In Tv line ab is // to xy hence its other view ab is TL. So draw x1y1 perpendicular to it.

3.Project view on x1y1.a) First draw projectors from ab & c on x1y1.b) from xy take distances of a,b & c( Tv) mark on these projectors from x1y1. Name points a1b1 & c1.c) This line view is an Aux.Tv. Draw x2y2 // to this line view and project Aux. Fv on it.

for that from x1y1take distances of ab & c and mark from x2y= on new projectors.4.Name points a1b1& c1 and join them. This will be the required true shape.

ALWAYS FOR NEW FV TAKEDISTANCES OF PREVIOUS FVAND FOR NEW TV, DISTANCES

OF PREVIOUS TV

REMEMBER!!

Problem 15: Fv & Tv of a triangular plate are shown.Determine its true shape.


84/159

x1

y1

c1

b1

a1

x2

y2

b1

c1

d1

c

X Y

a

b

b

ca

10

20

15

15

1

140

50

25USE SAME PROCEDURE STEPS

OF PREVIOUS PROBLEM:

BUT THERE IS ONE DIFFICULTY:

NO LINE IS // TO XY IN ANY VIEW.

MEANS NO TL IS AVAILABLE.

IN SUCH CASES DRAW ONE LINE

// TO XY IN ANY VIEW & ITS OTHER

VIEW CAN BE CONSIDERED AS TL

FOR THE PURPOSE.

HERE a 1 line in Fv is drawn // to xy.

HENCE its Tv a-1 becomes TL.

THEN FOLLOW SAME STEPS AND

DETERMINE TRUE SHAPE.

(STUDY THE ILLUSTRATION)

ALWAYS FOR NEW FV TAKEDISTANCES OF PREVIOUS FVAND FOR NEW TV, DISTANCES

OF PREVIOUS TV

REMEMBER!!


85/159

y1

X2

X1

a1c1

d1

b1

c1

d1

b1

a1

y2

TRUE SHAPEa

b

c

d YX

a

d

c

b

50 D.

50D

TL

PROBLEM 16:Fv & Tv both are circles of 50 mm diameter. Determine true shape of an elliptical plate.

ADOPT SAME PROCEDURE.a c is considered as line // to xy.Then ac becomes TL for the purpose.Using steps properly true shape can beEasily determined.

Study the illustration.

ALWAYS, FOR NEW FVTAKE DISTANCES OF

PREVIOUS FV ANDFOR NEW TV, DISTANCES

OF PREVIOUS TV

REMEMBER!!

Problem 17 : Draw a regular pentagon of30 mm sides with one side 300 inclined to xy


86/159

a

bc

d

e

a

be

cd

a1

b1

e1 d1

c1

300X Y

X1

Y1

450

30 mm sides with one side 30 inclined to xy.

This figure is Tv of some plane whose Fv is

A line 450 inclined to xy.

Determine its true shape.

IN THIS CASE ALSO TRUE LENGTH

IS NOT AVAILABLE IN ANY VIEW.

BUT ACTUALLY WE DONOT REQUIRE

TL TO FIND ITS TRUE SHAPE, AS ONE

VIEW (FV) IS ALREADY A LINE VIEW.

SO JUST BY DRAWING X1Y1 // TO THISVIEW WE CAN PROJECT VIEW ON IT

AND GET TRUE SHAPE:

STUDY THE ILLUSTRATION..

ALWAYS FOR NEW FVTAKE DISTANCES OF

PREVIOUS FV AND FORNEW TV, DISTANCES OF

PREVIOUS TV

REMEMBER!!

To understand and remember various solids in this subject properly


87/159

To understand and remember various solids in this subject properly,

those are classified & arranged in to two major groups.

Group A

Solids having top and base of same shape

Cylinder

Prisms

Triangular Square Pentagonal Hexagonal

Cube

Triangular Square Pentagonal Hexagonal

Cone

Tetrahedron

Pyramids

( A solid having

six square faces)( A solid having

Four triangular faces)

Group B

Solids having base of some shapeand just a point as a top, called apex.

SOLIDSDimensional parameters of different solids


88/159

Dimensional parameters of different solids.

Top

Rectangular

Face

Longer

Edge

Base

Edge

of

Base

Corner of

base

Corner of

base

Triangular

Face

Slant

Edge

Base

Apex

Square Prism Square Pyramid Cylinder Cone

Edge

of

Base

Base

Apex

Base

Generators

Imaginary lines

generating curved surface

of cylinder & cone.

Sections of solids( top & base not parallel) Frustum of cone & pyramids.

( top & base parallel to each other)

STANDING ON H.P RESTING ON H.P LYING ON H.P


89/159

X Y

STANDING ON H.P

On its base.

RESTING ON H.P

On one point of base circle.

LYING ON H.P

On one generator.

(Axis perpendicular to Hp

And // to Vp.)

(Axis inclined to Hp

And // to Vp)

(Axis inclined to Hp

And // to Vp)

While observing Fv, x-y line represents Horizontal Plane. (Hp)

Axis perpendicular to Vp

And // to Hp

Axis inclined to Vp

And // to Hp

Axis inclined to Vp

And // to Hp

X Y

F.V. F.V. F.V.

T.V. T.V. T.V.

While observing Tv, x-y line represents Vertical Plane. (Vp)

STANDING ON V.P

On its base.

RESTING ON V.P

On one point of base circle.

LYING ON V.P

On one generator.

STEPS TO SOLVE PROBLEMS IN SOLIDSProblem is solved in three steps:

STEP 1: ASSUME SOLID STANDING ON THE PLANE WITH WHICH IT IS MAKING INCLINATION.


90/159

( IF IT IS INCLINED TO HP, ASSUME IT STANDING ON HP)

( IF IT IS INCLINED TO VP, ASSUME IT STANDING ON VP)

IF STANDING ON HP - ITS TV WILL BE TRUE SHAPE OF ITS BASE OR TOP:

IF STANDING ON VP - ITS FV WILL BE TRUE SHAPE OF ITS BASE OR TOP.

BEGIN WITH THIS VIEW:ITS OTHER VIEW WILL BE A RECTANGLE ( IF SOLID IS CYLINDER OR ONE OF THE PRISMS):

ITS OTHER VIEW WILL BE A TRIANGLE ( IF SOLID IS CONE OR ONE OF THE PYRAMIDS):

DRAW FV & TV OF THAT SOLID IN STANDING POSITION:

STEP 2:CONSIDERING SOLIDS INCLINATION ( AXIS POSITION ) DRAW ITS FV & TV.

STEP 3:IN LAST STEP, CONSIDERING REMAINING INCLINATION, DRAW ITS FINAL FV & TV.

AXIS

VERTICAL

AXIS

INCLINED HP

AXISINCLINED VP

AXIS

VERTICAL

AXIS

INCLINED HP

AXIS

INCLINED VP

AXIS TO VPer

AXIS

INCLINED

VP

AXIS

INCLINED HP

AXIS TO VPer AXIS

INCLINED

VP

AXISINCLINED HP

GENERAL PATTERN ( THREE STEPS ) OF SOLUTION:

GROUP B SOLID.

CONE

GROUP A SOLID.

CYLINDER

GROUP B SOLID.

CONE

GROUP A SOLID.CYLINDER

Three steps

If solid is inclined to Hp

Three steps

If solid is inclined to Hp

Three steps

If solid is inclined to Vp

Study Next Twelve Problems and Practice them separately !!

Three steps

If solid is inclined to Vp

CATEGORIES OF ILLUSTRATED PROBLEMS!


91/159

PROBLEM NO.1, 2, 3, 4 GENERAL CASES OF SOLIDS INCLINED TO HP & VP

PROBLEM NO. 5 & 6 CASES OF CUBE & TETRAHEDRON

PROBLEM NO. 7 CASE OF FREELY SUSPENDED SOLID WITH SIDE VIEW.

PROBLEM NO. 8 CASE OF CUBE ( WITH SIDE VIEW)

PROBLEM NO. 9 CASE OF TRUE LENGTH INCLINATION WITH HP & VP.

PROBLEM NO. 10 & 11 CASES OF COMPOSITE SOLIDS. (AUXILIARY PLANE)

PROBLEM NO. 12 CASE OF A FRUSTUM (AUXILIARY PLANE)

CATEGORIES OF ILLUSTRATED PROBLEMS!

Problem 1. A square pyramid, 40mm base sides and axis 60 mm long,

has a triangular face on the ground

Solution Steps :Triangular face on Hp , means it is lying on Hp:

1.Assume it standing on Hp.

2.Its Tv will show True Shape of base( square)


92/159

XY

a

b c

d

o

o

dcba

o1

d1

b1c1

a1

a1

d1 c1

b1

o1

a1

(APEX

NEARER

TO V.P).(APEX

AWAY

FROM V.P.)

has a triangular face on the ground

and the vertical plane containing the

axis makes an angle of 450 with the

VP. Draw its projections. Take apex

nearer to VP

( )

3.Draw square of 40mm sides with one side vertical Tv &

taking 50 mm axis project Fv. ( a triangle)

4.Name all points as shown in illustration.

5.Draw 2ndFv in lying position I.e.ocd face on xy. And project its Tv.

6.Make visible lines dark and hidden dotted, as per the procedure.

7.Then construct remaining inclination with Vp( Vp containing axis ic the center line of 2nd Tv.Make it 450 to xy as

shown take apex near to xy, as it is nearer to Vp) & project final Fv.

For dark and dotted lines1.Draw proper outline of new view DARK. 2. Decide direction of an observer.

3. Select nearest point to observer and draw all lines starting from it-dark.

4. Select farthest point to observer and draw all lines (remaining)from it- dotted.

Problem 2:A cone 40 mm diameter and 50 mm axis

is resting on one generator on Hp

Solution Steps:Resting on Hp on one generator, means lying on Hp:

1.Assume it standing on Hp.

2 Its Tv will show True Shape of base( circle )


93/159

is resting on one generator on Hp

which makes 300 inclination with Vp


h

a

b

c

d

e

g

f

X Ya b dec g

fh

o

o

a1

h1

g1

f1

e1

d1

c1

b1

a1

c1

b1

d1

e1

f1

g1 h1

o1

a1b1

c1d1e1

f1

g1

h1

o1

o1

30

2.It s Tv will show True Shape of base( circle )

3.Draw 40mm dia. Circle as Tv &

taking 50 mm axis project Fv. ( a triangle)


5.Draw 2ndFv in lying position I.e.oe on xy. And

project its Tv below xy.

6.Make visible lines dark and hidden dotted,

as per the procedure.

7.Then construct remaining inclination with Vp

( generator o1e1 300 to xy as shown) & project final Fv.

For dark and dotted lines1.Draw proper outline of new vie

DARK.

2. Decide direction of an observer.

3. Select nearest point to observer

and draw all lines starting from

it-dark.

4. Select farthest point to observer

and draw all lines (remaining)from it- dotted.

Problem 3:A cylinder 40 mm diameter and 50 mm

i i ti i t f b

Solution Steps:Resting on Vp on one point of base, means inclined to Vp:

1.Assume it standing on Vp

2.Its Fv will show True Shape of base & top( circle )


94/159

X Ya b d c

1 2 4 3

a

b

c

d

1

2

3

4

450

4

3

2

1

d

c

b

a

350

a1

b1

c1d1

1

2

3

4

axis is resting on one point of a base

circle on Vp while its axis makes 450

with Vp and Fv of the axis 350 with Hp.

Draw projections..

2.It s Fv will show True Shape of base & top( circle )

3.Draw 40mm dia. Circle as Fv & taking 50 mm axis project Tv.

( a Rectangle)


5.Draw 2nd Tv making axis 450to xy And project its Fv above xy.

6.Make visible lines dark and hidden dotted, as per the procedure.7.Then construct remaining inclination with Hp

( Fv of axis I.e. center line of view to xy as shown) & project final Tv.

Problem 4:A square pyramid 30 mm base side

and 50 mm long axis is resting on its apex on Hp

Solution Steps :1.Assume it standing on Hp but as said on apex.( inverted ).

2.Its Tv will show True Shape of base( square)

3.Draw a corner case square of 30 mm sides as Tv(as shown)


95/159

b b1

X Y

a

d

co

d cba

o

c1a1

d1

o1

o1

a1b1

c1

d1

and 50 mm long axis is resting on its apex on Hp,

such that its one slant edge is vertical and a

triangular face through it is perpendicular to Vp.


q ( )

Showing all slant edges dotted, as those will not be visible from top.

4.taking 50 mm axis project Fv. ( a triangle)


6.Draw 2ndFv keeping oa slant edge vertical & project its Tv

7.Make visible lines dark and hidden dotted, as per the procedure.8.Then redrew 2nd Tv as final Tv keeping a1o1d1 triangular face

perpendicular to Vp I.e.xy. Then as usual project final Fv.

Problem 5: A cube of 50 mm long

edges is so placed on Hp on one

corner that a body diagonal is

Solution Steps:1.Assuming standing on Hp, begin with Tv,a square with all sides

equally inclined to xy.Project Fv and name all points of FV & TV.

2 D b d di l j i i ith 3( Thi b // t )


96/159

corner that a body diagonal is

parallel to Hp and perpendicular to

Vp Draw its projections.

X Y

b

c

d

a

a d cb

a1

b1

d1

c1

1

a1

d1

c1

d1

2.Draw a body-diagonal joining c with 3( This can become // to xy)

3.From 1 drop a perpendicular on this and name it p4.Draw 2ndFv in which 1-p line is vertical means c-3 diagonal

must be horizontal. .Now as usual project Tv..

6.In final Tv draw same diagonal is perpendicular to Vp as said in problem.Then as usual project final FV.

13 1

3

Problem 6:A tetrahedron of 50 mmlong edges is resting on one edge on

Hp while one triangular face containing

Solution StepsAs it is resting assume it standing on Hp.

Begin with Tv , an equilateral triangle as side case as shown:


97/159

Y

this edge is vertical and 450 inclined to

Vp. Draw projections.

X

T L

a o

b

c

ba c

o

a1

c1

o1

b1

900

450c1

a1

o1

b1

IMPORTANT:

Tetrahedron is aspecial type

of triangular

pyramid in which

base sides &

slant edges are

equal in length.

Solid of four faces.

Like cube it is also

described by One

dimension only..

Axis length

generally not given.

First project base points of Fv on xy, name those & axis line.

From a with TL of edge, 50 mm, cut on axis line & mark o

(as axis is not known, o is finalized by slant edge length)

Then complete Fv.

In 2ndFv make face obc vertical as said in problem.

And like all previous problems solve completely.


98/159

FREELY SUSPENDED SOLIDS:Positions of CG, on axis, from base, for different solids are shown below.

H

H/2

H/4

GROUP A SOLIDS

( Cylinder & Prisms)

GROUP B SOLIDS

( Cone & Pyramids)

CG

CG

Problem 7: A pentagonal pyramid

30 mm base sides & 60 mm long axis,

is freely suspended from one corner of

Solution Steps:In all suspended cases axis shows inclination with Hp.

1.Hence assuming it standing on Hp, drew Tv - a regular pentagon,corner case.

2 Project Fv & locate CG position on axis ( H from base ) and name g and


99/159

XY

a decb

o

a

bc

d

e

o

g

H/4

H

LINEdg VERTICAL

ab

c

d

o

e

g

a1

b1

o1

e1

d1

c1

a

e

d

c

b

FOR SIDE VIEW

is freely suspended from one corner of

base so that a plane containing its axis

remains parallel to Vp.

Draw its three views.

IMPORTANT:When a solid is freely

suspended from a

corner, then line

joining point of

contact & C.G.remains vertical.

( Here axis shows

inclination with Hp.)

So in all such cases,

assume solid standing

on Hp initially.)

2.Project Fv & locate CG position on axis( H from base.) and name g and

Join it with corner d

3.As 2nd Fv, redraw first keeping line gd vertical.

4.As usual project corresponding Tv and then Side View looking from.

Problem 8:A cube of 50 mm long edges is so placed

on Hp on one corner that a body diagonal

th h thi i di l t H

Solution Steps:1.Assuming it standing on Hp begin with Tv, a square of corner case.

2.Project corresponding Fv.& name all points as usual in both views.

3 Join a1 as body diagonal and draw 2nd Fv making it vertical (I on xy)


100/159

a d cb

b

c

d

a a1

b

d1

c1

d

c

a

b

X Y1 11

through this corner is perpendicular to Hp

and parallel to Vp Draw its three views.

3.Join a 1 as body diagonal and draw 2 Fv making it vertical (I on xy)

4.Project its Tv drawing dark and dotted lines as per the procedure.

5.With standard method construct Left-hand side view.( Draw a 450 inclined Line in Tv region ( below xy).

Project horizontally all points of Tv on this line and

reflect vertically upward, above xy.After this, drawhorizontal lines, from all points of Fv, to meet these

lines. Name points of intersections and join properly.

For dark & dotted lines

locate observer on left side of Fv as shown.)

Problem 9: A right circular cone,40 mm base diameter and 60 mm

long axis is resting on Hp on one

point of base circle such that its

This case resembles to problem no.7 & 9 from projections of planes topic.In previous all cases 2nd inclination was done by a parameter not showing TL.Like

Tv of axis is inclined to Vp etc. But here it is clearly said that the axis is 400 inclined

to Vp Means here TL inclination is expected So the same construction done in those


101/159

1

400

Axis Tv Length

Axis Tv Length

Axis True Length

Locus of

Center 1

c1

a1

b1

e1

d1

h1

f1

g1

o1

h

a

b

c

d

e

g

f

yX a b decg fh

o

450

a1

h1 f1

e1

d1

c1

b1

g1

o11

point of base circle such that it s

axis makes 450 inclination with

Hp and 400 inclination with Vp.


to Vp. Means here TL inclination is expected. So the same construction done in those

Problems is done here also. See carefully the final Tv and inclination taken there.

So assuming it standing on HP begin as usual.

Problem 10: A triangular prism,

40 mm base side 60 mm axis

is lying on Hp on one rectangular face


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450

F.V.

T.V.

Aux.F.V.

X Y

with axis perpendicular to Vp.

One square pyramid is leaning on its face

centrally with axis // to vp. Its base side is

30 mm & axis is 60 mm long resting on Hp

on one edge of base.Draw FV & TV ofboth solids.Project another FV

on an AVP 450 inclined to VP.

Steps :

Draw Fv of lying prism

( an equilateral Triangle)

And Fv of a leaning pyramid.Project Tv of both solids.

Draw x1y1 450 inclined to xy

and project aux.Fv on it.

Mark the distances of first FV

from first xy for the distances

of aux. Fv from x1y1 line.

Note the observers directions

Shown by arrows and furthersteps carefully.

Problem 11:A hexagonal prism ofbase side 30 mm longand axis 40 mm long,is standing on Hp on its base with

one base edge // to Vp.


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X Y

X1

Y1

o

o

Fv

Tv

Aux.Tv

450

A tetrahedron is placed centrally

on the top of it.The base of tetrahedron is

a triangle formed by joining alternate corners

of top of prism..Draw projections of both solids.

Project an auxiliary Tv on AIP 450

inclined to Hp.

a b dc ef

a

b c

d

ef

STEPS:

Draw a regular hexagon as Tv of

standing prism With one side // to xy

and name the top points.Project its Fv

a rectangle and name its top.

Now join its alternate corners

a-c-e and the triangle formed is baseof a tetrahedron as said.

Locate center of this triangle

& locate apex o

Extending its axis line upward

mark apex o

By cutting TL of edge of tetrahedron

equal to a-c. and complete Fvof tetrahedron.

Draw an AIP ( x1y1) 450 inclined to xy

And project Aux.Tv on it by using similar

Steps like previous problem.

a1

b1

c1

d1

e1

f1

o1

Problem 12:A frustum of regular hexagonal pyramid is standing on its larger baseOn Hp with one base side perpendicular to Vp.Draw its Fv & Tv.

Project its Aux.Tv on an AIP parallel to one of the slant edges showing TL.

Base side is 50 mm long , top side is 30 mm long and 50 mm is height of frustum.


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X Y

X1

Y1 AIP // to slant edgeShowing true lengthi.e. a- 1

a b e c d

1 25 34

Fv

Tv

Aux.Tv

1

23

45

a

b

d

c

e

1 2

34

5

b1

c1d1

e1

a1

g , p g g

ENGINEERING AP

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